\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt[3]{{1 - {x^3}}}}}{{4x + 1}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt[3]{{1 - {x^3}}}}}{{4x + 1}} = \mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt[3]{{\dfrac{1}{{{x^3}}} - 1}}}}{{4 + \dfrac{1}{x}}} = \dfrac{{ - 1}}{4}\).
Chọn B.